Question

an observer stands 200 meters from the launch site of a hot-air balloon. balloon rises 4m/sec at a constant rate. how fast is the angle of elevation of the balloon increasing 30sec after the launch?

Answer 1

after 30sec, the balloon reaches a height of 120m.
the angle is then:
\[\alpha=\tan^{-1} (120/200) = 31degrees\]

Answer 2

1.36

Answer 3

sorry about the 1.36- I had a bad connection and couldnt finish. \[d \theta/dt=4m/(200\sec ^{2}\theta)\] and \[\sec ^{2}\theta=1.36\]